Tr`es bien. Then I can find a by taking cube roots. I also can find b in the same way, or by using b = m/3a. Therefore, x = a − b is the solution to the cubic equation. Am I correct? Well, I substitute x into the original cubic equation and see if it’s true. I could ask Antoine-August to show my analysis to his tutor. He would tell me if I’m doing this correctly. Friday | September 25, 1789 Autumn is slowly making an entrance. The weather is so nice, and the colors of the foliage are changing. After lunch my sisters and I went for a stroll to the Jardin des Tuileries.

Master” — 2012/3/8 — 13:19 — page 35 — #47 ✐ ✐ 1. Awakening 35 What am I missing? I will not sleep well if I do not solve it. I will attempt a different approach. First I see that (a − b)3 + 3ab(a − b) = a3 − b3 . Then, if a and b satisfy 3ab = m, and a3 − b3 = n, then a − b is a solution of x3 + mx = n. Now b = m/3a, so a3 − m3 /27a3 = n, which I can also write as a6 − na3 − m3 /27 = 0. This is like a quadratic equation in a3 , (a3 )2 − n(a3 ) − (m3 )/27 = 0. So, I can solve for a3 using the formula for a quadratic equation.

Is this correct? Oh, how can I be certain? If a polynomial is a mathematical expression involving a series of powers in one or more variables multiplied by coefficients, then I can write a polynomial in one variable with constant coefficients as: an xn + · · · + a3 x3 + a2 x2 + a1 x + a0 = 0. The highest power in the polynomial is called its order. And it makes more sense if I write the polynomial as: x3 + 2x2 + 3x + 5 = 0. ✐ ✐ ✐ ✐ ✐ ✐ “master” — 2012/3/8 — 13:19 — page 20 — #32 ✐ ✐ 20 Sophie s' Diary Then I say that this is a third-order polynomial.